Assume that \(\Phi _{ef}^{\ast }(p)=1\) . Since \(n\geqslant 3\)  by Remark REF ,
it follows that \((ef,p)=(6,2)\)  (see above). Then \(X\)  is isomorphic to one of
the groups \(PSL_{3}(4)\) , \(PSU_{4}(2)\) , \(PSL_{6}(2)\) , \(PSp_{6}(2)\) , \(P\Omega _{6}^{-}(2)\)  or \(P\Omega _{8}^{+}(2)\) .
Assume that \(X\cong PSL_{3}(4)\) . Then \(35\mid \left|X_{x}\right|\) ,
since \(\left[ X:X_{x}\right] =k^{2}\) , but \(X\)  does not contain such a group
by [1]}.
Assume that \(X\cong PSU_{4}(2)\) . Then \(PSU_{4}(2)\trianglelefteq G\leqslant PSU_{4}(2):Z_{2}\) . Also \(5\mid \left|X_{x}\right|\)  and \(\left[X:X_{x}\right] =k^{2}\) , with \(k\mid 72\)  and \(k+1\mid \left|G\right|\) . Easy computations show that \(k=3,4,8\)  or 9. However, \(X\)  does not have \(k^{2}\)  as a transitive permutation of degree for any such values of \(k\)  by [1]}.
Assume that \(X\cong PSL_{6}(2)\) . Thus \(2\cdot 5\cdot 31\mid \left|X_{x}\right|\) , since \(\left[ X:X_{x}\right] =k^{2}\) , and hence either \(X_{x}\cong SL_{5}(2)\)  or \(X_{x}\cong E_{2^{5}}:SL_{5}(2)\)  by [3]},
Tables 8.24 and 8.25, and by [1]}. Since \(\left|X_{x}\right|_{7}=7\)  and \(\left|X \right|_{7}=7^{2}\) , it follows that \(\left[ X:X_{x}\right]_{7}=7 \) , but this contradicts \(\left[X:X_{x}\right] =k^{2}\) .
Assume that \(X\cong PSp_{6}(2)\) . Then \(G=X\) , since \(\mathrm {Out(}X\mathrm {)}=1\) . Also \(\left[ G:G_{x}\right] =k^{2}\)  implies \(2\cdot 5\cdot 7\mid \left|G_{x}\right|\) . Then either \(G_{x}=A_{7}\)  and \(k=24\) , or \(G_{x}\cong S_{8}\)  and \(k=6\)  by [1]}. The former case is ruled out,
since it contradicts Lemma REF (3). Indeed, \(k+1=5^{2}\)  does not divide \(\left|G_{x}\right|\) . Thus, \(G_{x}\cong S_{8}\) , \(k=6\)  and hence \(\lambda =2,3\)  or 6, as \(\lambda \mid k\)  and \(\lambda \geqslant 2\) . However, \(S_{8}\)  does not have transitive permutation representations of degree
\(r=14,21\)  or 42 by [1]}. Thus \(X\cong PSp_{6}(2)\)  is
excluded.
Assume that \(X\cong P\Omega _{6}^{-}(2)\) . Then \(5\mid \left|X_{x}\right|\) , since \(\left[ X:X_{x}\right] =k^{2}\) , and hence \(k\mid 2^{3}\cdot 3^{2}\) . Also \(\frac{k+1}{(k+1,2)}\)  divides \(\left|X\right|\)  by Lemma REF , since \(\mathrm {Out(}X\mathrm {)}\cong Z_{2}\) . Easy computations show that \(k=3,4,8\)  or 9. However, for these \(k\) 
the group \(X\)  does not have a transitive permutation representation of
degree \(k^{2}\)  by [1]}, and so \(X\cong P\Omega _{6}^{-}(2)\)  is
ruled out too.
Assume that \(X\cong P\Omega _{8}^{+}(2)\) . Therefore \(21\mid \left|X_{x}\right|\) , since \(\left[ X:X_{x}\right] =k^{2}\) , and hence \(k^{2}\mid 2^{12}\cdot 3^{4}\cdot 5^{2}\) . Also \(\left[ X:X_{x}\right] \)  must
be divisible by 120, 135 or 960 by [1]}. Then 15 divides \(\left[X:X_{x}\right] \)  in each of these cases. Actually, \(15\mid k\)  as \(\left[X:X_{x}\right] =k^{2}\) . Then \(k=15\theta \) , with \(\theta \mid 2^{6}\cdot 3\) 
and \(\theta >1\) , and \(\frac{k+1}{(k+1,2)}\mid \left|X\right|\)  as \(\mathrm {Out(}X\mathrm {)}\cong S_{3}\) . It is straightforward to check that \(\frac{k+1}{(k+1,2)}\nmid \left|X\right|\)  for any \(\theta \mid 2^{6}\cdot 3\)  and \(\theta >1\) , and hence \(X\cong P\Omega _{8}^{+}(2)\)  cannot
occur.
